anyone good w/ rate law stuff?

[springserene]

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-----------[A]initial ------ initial ------ rate
1)-------- 1M -------------- 1M ------------ 2
2)-------- 1M -------------- 3M ------------ 18

Shouldn't B be 3rd order b/c

(1/3)(18/2) = 3

kaplan says B is 2nd order b/c tripling results in 3x3=9 multilple of rate. i'm confused.

 

[SteakNature]

The rate goes from 2 to 18. Therefore, the rate is multiplied by 9.

Since the concentration of A remains constant when you go from reaction 1 (rate is 2) to reaction 2 (rate is 18), you look for the concentration of B.

The concentration of B is tripled.

The only way you can multiply the rate by 9, while tripling B and having A constant is to have the concentration of B squared.
B is then second ordered.

Don't forget that the formula for rates is : rate = k [A]^x ^y (^ meaning power)

Hope that helps.

 

[springserene]

thx!!!
i understand your reasoning.